integral of sin^2(1-2x)

الحلّ

\int sin^{2} (1 - 2 x) d x

\frac{1}{2} (x + \frac{1}{4} sin (2 - 4 x)) + C

خطوات الحلّ

\int sin^{2} (1 - 2 x) d x

Rewrite using trig identities

= \int \frac{1 - cos ( 2 ( 1 - 2 x ) )}{2} d x

\int a \cdot f (x) d x = a \cdot \int f (x) d x

:قم بإخراج الثابت

= \frac{1}{2} \cdot \int 1 - cos (2 (1 - 2 x)) d x

2 (1 - 2 x)

وسٌع:

2 - 4 x

= \frac{1}{2} \cdot \int 1 - cos (2 - 4 x) d x

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

:فعّل الميزة الخطيّة

= \frac{1}{2} (\int 1 d x - \int cos (2 - 4 x) d x)

\int 1 d x = x

\int cos (2 - 4 x) d x = - \frac{1}{4} sin (2 - 4 x)

= \frac{1}{2} (x - (- \frac{1}{4} sin (2 - 4 x)))

بسّط

= \frac{1}{2} (x + \frac{1}{4} sin (2 - 4 x))

أضف ثابت التكامل للحل

= \frac{1}{2} (x + \frac{1}{4} sin (2 - 4 x)) + C

\int \frac{x ^{2} + 2 x + 3}{( x + 1 ) ( x ^{2} + 2 x + 4 )} d x

\int θ tan (3) θ^{2} d θ

\int \frac{cos ( 2 x )}{sin ^{5} ( 2 x )} d x

\int 16 x^{2} - x^{4} d x

\int - x ln (x^{2} + 1) d x